A short bar magnet placed with its axis at 30° with an external field of 800 Gauss, experiences a torque of 0.016 N.m. The work done in moving it from most stable to most unstable position is $$\alpha \times 10^{-3}J$$. The value of $$\alpha$$ is ___.

We need to find the work done in moving a short bar magnet from the most stable to the most unstable position.

Angle with external field: $$\theta = 30°$$

External field: $$B = 800$$ Gauss = 0.08 T

Torque: $$\tau = 0.016$$ N.m

Torque on a magnetic dipole: $$\tau = mB\sin\theta$$

$$0.016 = m \times 0.08 \times \sin 30°$$

$$0.016 = m \times 0.08 \times 0.5 = 0.04m$$

$$m = \frac{0.016}{0.04} = 0.4 \text{ A.m}^2$$

The most stable position is when $$\theta = 0°$$ (magnet aligned with the field).

The most unstable position is when $$\theta = 180°$$ (magnet anti-parallel to the field).

Work done is the change in potential energy:

$$W = U_{final} - U_{initial} = (-mB\cos 180°) - (-mB\cos 0°)$$

$$W = mB + mB = 2mB$$

$$W = 2 \times 0.4 \times 0.08 = 0.064 \text{ J} = 64 \times 10^{-3} \text{ J}$$

Therefore, $$\alpha = $$ 64.